scalike-jdbc - 如何在查询中指定列别名?

scalike-jdbc - How to specify column alias in query?

我定义了一个对象来帮助将结果集转换为 POJO。

  // an object, that help to map from query result to POJO,
  object Dummy extends SQLSyntaxSupport[Dummy] {
    override val tableName = "dummy"
    def apply(rs: WrappedResultSet) = new Dummy(
      rs.long("id"), rs.stringOpt("name"), rs.intOpt("size"), rs.jodaDateTime("create_date"))
  }

然后下面的代码尝试从table,

查询
  /*** query - with condition - start ***/
  // use paste mode (:paste) on the Scala REPL
  val d = Dummy.syntax("d")
  val name = "hello"
  val helloDummyOpt: Option[Dummy] = withSQL {
    select.from(Dummy as d).where.eq(d.name, name).limit(1)
  }.map(rs => Dummy(rs)).single.apply()

  printf("hello dummy:\t%s\n", if(helloDummyOpt.isDefined) helloDummyOpt.get else "")
  /*** query - with condition - end ***/

从控制台,我可以看到生成的 sql 是:

select d.id as i_on_d, d.name as n_on_d, d.size as s_on_d, d.create_date as cd_on_d
from dummy d where d.name = 'hello' limit 1;

为了使转换工作,需要更改生成的 sql 中的列别名以保留 table 中的原始名称,例如 name 而不是 n_on_d.

那么,如何做到这一点,谢谢。


@更新

完整代码在这里:

package eric.scalikejdbc

import scalikejdbc._
import org.postgresql.Driver._
import org.joda.time._

/**
 * scalikejdbc hello.
 * @author eric
 * @date Jul 6, 2016 23:14:07 PM
 */
object ScalikeJdbcHello extends App {
  Class.forName("org.postgresql.Driver")
  ConnectionPool.singleton("jdbc:postgresql://localhost:5432/sportslight", "postgres", "123456")

  implicit val session = AutoSession

  /*** query - parse as List of Map - start ***/
  // find rows, with limit 10000,
  val entities: List[Map[String, Any]] = sql"select * from dummy limit 10000".map(_.toMap).list.apply()
  printf("dummy count: %d\n", entities.size)
  /*** query - parse as List of Map - end ***/

  /*** query - parse as POJO - start ***/
  // define a POJO class,
  case class Dummy(id: Long, name: Option[String], size: Option[Int], createDate: DateTime)

  // an object, that help to map from query result to POJO,
  object Dummy extends SQLSyntaxSupport[Dummy] {
    override val tableName = "dummy"
    def apply(rs: WrappedResultSet) = new Dummy(
      rs.long("id"), rs.stringOpt("name"), rs.intOpt("size"), rs.jodaDateTime("create_date"))
  }

  // find top 10 rows,
  val dummyList: List[Dummy] = sql"select * from dummy limit 10".map(rs => Dummy(rs)).list.apply()

  var i = 0;
  printf("dummy top rows:\n")
  for (d <- dummyList) {
    printf("\t[%d] %s\n", i, d)
    i = i + 1
  }
  /*** query - parse as POJO - end ***/

  /*** query - with condition - start ***/
  // TODO ... specify column name in result,
  val d = Dummy.syntax("d")
  val name = "hello"
  val helloDummyOpt: Option[Dummy] = withSQL {
    select.from(Dummy as d).where.eq(d.name, name).limit(1)
  }.map(rs => Dummy(rs)).single.apply()

  printf("hello dummy:\t%s\n", if(helloDummyOpt.isDefined) helloDummyOpt.get else "")
  /*** query - with condition - end ***/
}

这里是 sql 创建 table 并为 postgresql 初始化数据:

/* table - dummy */
-- drop table
drop table if exists dummy;

-- create table
create table dummy (
    id serial8,
    name varchar(50) not null,
    size int not null,
    create_date timestamptz,
    primary key (id)
);

-- init data - dummy
insert into dummy(name, size, create_date) values
('test', 1, now()),
('hello', 1, now());

您使用的是半类型安全 DSL,半默认 API。您应该选择一个并继续。

对于 TypeSafe DSL,您不应使用自定义 SqlSyntaxSupport 实现。

而不是 .map(rs => Dummy(rs)) 您应该能够使用 d Dummy 对象进行操作,该对象在 sql 查询中定义。

  val d = Dummy.syntax("d")

  val helloDummyOpt: Option[Dummy] = withSQL {
    select.from(Dummy as d).where.eq(d.name, name).limit(1)
  }.single.apply()

查看 TypeSafe DSL 部分以获取更多示例:http://scalikejdbc.org

如果您想接收那些自动生成的名称,您应该通过对象 d:

引用它们
 val d = Dummy.syntax("d")
d.id // = i_on_d
d.name // = n_on_d
d.create_date // = cd_on_d

select.from 班级 alias.result.*。调用 alias.* 按原样简化 returns 列名称。

scala> select(a.*).from(Article as a).toSQL.statement
res7: String = select a.id, a.title, a.body, a.created_at, a.updated_at from articles a

scala> select(a.result.*).from(Article as a).toSQL.statement
res8: String = select a.id as i_on_a, a.title as t_on_a, a.body as b_on_a, a.created_at as ca_on_a, a.updated_at as ua_on_a from articles a

scala> select.from(Article as a).toSQL.statement
res9: String = select a.id as i_on_a, a.title as t_on_a, a.body as b_on_a, a.created_at as ca_on_a, a.updated_at as ua_on_a from articles a